1. The Unit Circle

The Unit Circle

The unit circle is one of the most important concepts in trigonometry. It connects angles, coordinates, right triangles, and trigonometric functions in a single mathematical model.

A unit circle is a circle with a radius of exactly 1 unit, centred at the origin (0, 0) of a Cartesian coordinate system.

The unit circle allows us to determine exact values of sine, cosine, and tangent for common angles. It also helps explain why trigonometric functions can have positive and negative values and why their values repeat.

Unlike right-triangle trigonometry, which initially focuses on acute angles, the unit circle allows us to work with angles of any size, including angles greater than 90°, negative angles, and angles greater than 360°.

Defining the Unit Circle

The equation of a circle centred at the origin with radius r is:

x² + y² = r²

For a unit circle:

r = 1

Therefore:

x² + y² = 1

Every point (x, y) on the unit circle satisfies this equation.

For example, consider the point:

(0.6, 0.8)

Check:

0.6² + 0.8² = 0.36 + 0.64 = 1

Therefore, (0.6, 0.8) lies on the unit circle.

The point (2, 1) does not lie on the unit circle because:

2² + 1² = 5

which is not equal to 1.

Important Features

The unit circle has:

• Centre: (0, 0)

• Radius: 1 unit

• Diameter: 2 units

• Circumference: 2π units

• Equation: x² + y² = 1

The circle crosses the coordinate axes at:

(1, 0), (0, 1), (−1, 0), and (0, −1).

These four points correspond to the quadrantal angles 0°, 90°, 180°, and 270°.

Angles in Standard Position

An angle is in standard position when:

• Its vertex is at the origin.

• Its initial side lies along the positive x-axis.

• Its terminal side is determined by rotating from the initial side.

Angles are measured from the positive x-axis.

Positive angles are measured counterclockwise.

Negative angles are measured clockwise.

For example:

• 30° represents a counterclockwise rotation of 30°.

• 90° represents a quarter-turn counterclockwise.

• 180° represents a half-turn.

• 270° represents three-quarters of a full turn.

• −90° represents a quarter-turn clockwise.

A complete revolution corresponds to 360°.

Angles differing by a whole number of complete revolutions have the same terminal side.

These are called coterminal angles.

For example:

30°, 390°, and −330°

all have the same terminal side.

Degrees and Radians

Angles can be measured in degrees or radians.

A complete circle contains:

360° = 2π radians

Therefore:

180° = π radians

90° = π/2 radians

45° = π/4 radians

30° = π/6 radians

60° = π/3 radians

Converting Degrees to Radians

Use:

Radians = Degrees × π/180

Example:

Convert 120° to radians.

120 × π/180 = 2π/3

Therefore:

120° = 2π/3 radians

Converting Radians to Degrees

Use:

Degrees = Radians × 180/π

Example:

Convert 5π/6 to degrees.

(5π/6) × (180/π) = 150

Therefore:

5π/6 radians = 150°

Common Angle Conversions

Radians are especially important in calculus because the standard derivative and integral formulas for trigonometric functions assume angles are measured in radians.

Sine and Cosine on the Unit Circle

The most important relationship on the unit circle is:

(x, y) = (cos θ, sin θ)

where θ is the angle measured from the positive x-axis.

This means:

cos θ = x-coordinate

sin θ = y-coordinate

Every angle corresponds to a point on the unit circle.

Example: 30°

At an angle of 30°, the point on the unit circle is:

(√3/2, 1/2)

Therefore:

cos 30° = √3/2

sin 30° = 1/2

Example: 90°

At 90°, the point is:

(0, 1)

Therefore:

cos 90° = 0

sin 90° = 1

Example: 180°

At 180°, the point is:

(−1, 0)

Therefore:

cos 180° = −1

sin 180° = 0

The unit circle makes it possible to determine sine and cosine directly from coordinates.

Why the Coordinates Equal Cosine and Sine

Consider a point P(x, y) on the unit circle in the first quadrant.

Draw a perpendicular line from P to the x-axis.

This creates a right triangle with:

• Hypotenuse = 1

• Horizontal side = x

• Vertical side = y

Using right-triangle trigonometry:

cos θ = adjacent/hypotenuse

Since the hypotenuse is 1:

cos θ = x/1

cos θ = x

Similarly:

sin θ = opposite/hypotenuse

sin θ = y/1

sin θ = y

This relationship extends to all quadrants using signed coordinates.

It also explains why:

sin² θ + cos² θ = 1

This identity follows directly from the equation of the unit circle.

Exact Values for Special Angles

Certain angles have exact trigonometric values that can be found using special right triangles.

The most important are:

• 0°

• 30°

• 45°

• 60°

• 90°

The 45°–45°–90° Triangle

An isosceles right triangle has two equal legs.

If each leg has length 1, the hypotenuse is:

√(1² + 1²) = √2

To create a triangle with hypotenuse 1, divide each side by √2.

The legs become:

1/√2 = √2/2

Therefore:

sin 45° = √2/2

cos 45° = √2/2

The 30°–60°–90° Triangle

An equilateral triangle with side length 2 can be divided into two right triangles.

Each right triangle has:

• Hypotenuse = 2

• Shorter leg = 1

• Longer leg = √3

Dividing all sides by 2 gives a hypotenuse of 1.

Therefore:

sin 30° = 1/2

cos 30° = √3/2

sin 60° = √3/2

cos 60° = 1/2

Exact Trigonometric Values in the First Quadrant

These values are fundamental to trigonometry.

Rather than memorising every angle around the circle separately, we can use these first-quadrant values together with quadrant signs.

The Four Quadrants

The coordinate plane is divided into four quadrants.

Quadrant I

Angles between 0° and 90°.

x is positive.

y is positive.

Therefore:

• cos θ is positive.

• sin θ is positive.

• tan θ is positive.

Quadrant II

Angles between 90° and 180°.

x is negative.

y is positive.

Therefore:

• cos θ is negative.

• sin θ is positive.

• tan θ is negative.

Quadrant III

Angles between 180° and 270°.

x is negative.

y is negative.

Therefore:

• cos θ is negative.

• sin θ is negative.

• tan θ is positive.

Quadrant IV

Angles between 270° and 360°.

x is positive.

y is negative.

Therefore:

• cos θ is positive.

• sin θ is negative.

• tan θ is negative.

Quadrant Sign Summary

A useful memory aid is ASTC:

• All positive in Quadrant I.

• Sine positive in Quadrant II.

• Tangent positive in Quadrant III.

• Cosine positive in Quadrant IV.

The quadrant determines the sign of a trigonometric value, while the reference angle determines its magnitude.

Reference Angles

A reference angle is the acute angle between the terminal side of an angle and the x-axis.

Reference angles allow us to use familiar first-quadrant values to evaluate trigonometric functions in any quadrant.

For angles between 0° and 360°:

For quadrantal angles, the reference angle is 0°, and the axis coordinates should be used directly.

Worked Example: 150°

Find sin 150° and cos 150°.

150° lies in Quadrant II.

Reference angle:

α = 180° − 150° = 30°

In Quadrant II:

• Sine is positive.

• Cosine is negative.

Therefore:

sin 150° = 1/2

cos 150° = −√3/2

Worked Example: 225°

225° lies in Quadrant III.

Reference angle:

α = 225° − 180° = 45°

In Quadrant III, sine and cosine are both negative.

Therefore:

sin 225° = −√2/2

cos 225° = −√2/2

Worked Example: 300°

300° lies in Quadrant IV.

Reference angle:

α = 360° − 300° = 60°

In Quadrant IV:

• Cosine is positive.

• Sine is negative.

Therefore:

cos 300° = 1/2

sin 300° = −√3/2

Coordinates Around the Entire Unit Circle

The following table gives the exact coordinates for the most commonly used angles.

Remember:

(x, y) = (cos θ, sin θ)

The exact values follow a repeating pattern of magnitudes and signs.

Tangent on the Unit Circle

Tangent is defined by:

tan θ = sin θ / cos θ

Since:

sin θ = y

and:

cos θ = x

we can also write:

tan θ = y/x

provided x ≠ 0.

Worked Example: 60°

sin 60° = √3/2

cos 60° = 1/2

Therefore:

tan 60° = (√3/2)/(1/2)

tan 60° = √3

Worked Example: 135°

sin 135° = √2/2

cos 135° = −√2/2

Therefore:

tan 135° = (√2/2)/(−√2/2)

tan 135° = −1

When Is Tangent Undefined?

Tangent is undefined when cosine equals zero.

This occurs at:

90°, 270°, and coterminal angles.

For example:

tan 90° = 1/0

Division by zero is undefined.

Other Trigonometric Functions

The unit circle can also be used to evaluate cosecant, secant, and cotangent.

These functions are reciprocals of sine, cosine, and tangent.

csc θ = 1/sin θ

sec θ = 1/cos θ

cot θ = cos θ/sin θ

Worked Example: 30°

We know:

sin 30° = 1/2

cos 30° = √3/2

Therefore:

csc 30° = 1/(1/2)

csc 30° = 2

For secant:

sec 30° = 1/(√3/2)

= 2/√3

sec 30° = 2√3/3

For cotangent:

cot 30° = (√3/2)/(1/2)

cot 30° = √3

Undefined Values

• Cosecant is undefined when sine is zero.

• Secant is undefined when cosine is zero.

• Cotangent is undefined when sine is zero.

These restrictions are important when evaluating trigonometric expressions.

Negative Angles

Negative angles are measured clockwise from the positive x-axis.

For example:

−30° is coterminal with 330°.

Therefore:

cos(−30°) = cos 330° = √3/2

sin(−30°) = sin 330° = −1/2

This illustrates two important identities:

cos(−θ) = cos θ

sin(−θ) = −sin θ

Cosine is an even function.

Sine is an odd function.

Tangent is also odd:

tan(−θ) = −tan θ

Worked Example

Evaluate:

sin(−45°)

Since:

sin 45° = √2/2

Therefore:

sin(−45°) = −√2/2

Angles Greater Than 360°

Angles greater than 360° represent more than one complete revolution.

Because a complete revolution is 360°, we can subtract multiples of 360° to find a coterminal angle.

Worked Example: 420°

420° − 360° = 60°

Therefore:

sin 420° = sin 60°

sin 420° = √3/2

Worked Example: 765°

765° − 720° = 45°

Therefore:

cos 765° = cos 45°

cos 765° = √2/2

In radians, the corresponding period for sine and cosine is 2π.

Thus:

sin(θ + 2π) = sin θ

cos(θ + 2π) = cos θ

Tangent repeats every π radians, or 180°.

Trigonometric Identities from the Unit Circle

The unit circle provides a geometric explanation for several important trigonometric identities.

The Pythagorean Identity

The equation of the unit circle is:

x² + y² = 1

Since:

x = cos θ

y = sin θ

we obtain:

cos² θ + sin² θ = 1

This is the fundamental Pythagorean trigonometric identity.

Rearranging the Identity

We can write:

sin² θ = 1 − cos² θ

or:

cos² θ = 1 − sin² θ

These relationships allow us to calculate one trigonometric value when another is known.

Worked Example

Suppose:

cos θ = 3/5

and θ lies in Quadrant IV.

Find sin θ.

Use:

sin² θ + cos² θ = 1

Substitute:

sin² θ + 9/25 = 1

sin² θ = 16/25

Therefore:

sin θ = ±4/5

Since θ lies in Quadrant IV, sine is negative.

sin θ = −4/5

Using the Unit Circle to Solve Trigonometric Equations

The unit circle can help us find angles that satisfy trigonometric equations.

Worked Example: sin θ = 1/2

Find all solutions between 0° and 360°, including 0° but excluding 360°.

The reference angle is 30°.

Sine is positive in Quadrants I and II.

Therefore:

θ = 30° or 150°

Worked Example: cos θ = −√2/2

The reference angle is 45°.

Cosine is negative in Quadrants II and III.

Therefore:

θ = 135° or 225°

Worked Example: tan θ = −√3

The reference angle is 60°.

Tangent is negative in Quadrants II and IV.

Therefore:

θ = 120° or 300°

These methods are important when solving trigonometric equations in algebra, calculus, and physics.

Connecting the Unit Circle to Trigonometric Graphs

As an angle increases around the unit circle, the x-coordinate and y-coordinate change continuously.

These changes produce the cosine and sine graphs.

Sine Graph

The sine function represents the y-coordinate of the rotating point.

Its values range from −1 to 1.

Its period is:

360° or 2π radians

Cosine Graph

The cosine function represents the x-coordinate.

Its values also range from −1 to 1.

Its period is:

360° or 2π radians

Tangent Graph

Tangent is the ratio of sine to cosine.

Its values are not restricted to the interval from −1 to 1.

The tangent function has vertical asymptotes where cosine is zero.

Its period is:

180° or π radians

These relationships explain the repeating patterns of trigonometric functions.

Applications of the Unit Circle

The unit circle is used in many areas of mathematics, science, and engineering.

Circular Motion

A point moving around a circle of radius r can be described by:

x = r cos θ

y = r sin θ

For a unit circle, r = 1.

If the angular position changes at a constant rate ω:

θ = ωt + θ₀

then:

x(t) = r cos(ωt + θ₀)

y(t) = r sin(ωt + θ₀)

These equations describe uniform circular motion.

Waves and Oscillations

Sine and cosine functions describe many repeating phenomena, including:

• Sound waves

• Water waves

• Alternating electrical current

• Pendulum motion for small oscillations

• Vibrations in mechanical systems

A simple oscillation may be modelled by:

y(t) = A sin(ωt + φ)

where:

• A = amplitude

• ω = angular frequency

• φ = phase constant

• t = time

Engineering and Navigation

Sine and cosine can be used to resolve a vector into horizontal and vertical components.

For a vector with magnitude F and direction θ measured from the positive x-axis:

Fₓ = F cos θ

Fᵧ = F sin θ

These equations are used in force analysis, navigation, projectile motion, and engineering design.

Worked Example: Force Components

A force of 100 N acts at an angle of 30° above the positive horizontal axis.

Horizontal component:

Fₓ = 100 cos 30°

Fₓ = 100(√3/2)

Fₓ = 50√3 N ≈ 86.6 N

Vertical component:

Fᵧ = 100 sin 30°

Fᵧ = 100(1/2)

Fᵧ = 50 N

The unit circle provides the exact trigonometric values needed for the calculation.

A Reliable Strategy for Unit Circle Problems

When evaluating a trigonometric function:

1. Identify the angle.

2. Convert between degrees and radians if necessary.

3. Reduce the angle to a coterminal angle between 0° and 360°.

4. Determine the quadrant.

5. Find the reference angle.

6. Recall the exact value for the reference angle.

7. Apply the correct sign.

8. Simplify the result.

9. Check whether the function is defined.

Example

Evaluate:

cos 510°

Subtract 360°:

510° − 360° = 150°

150° lies in Quadrant II.

Reference angle:

180° − 150° = 30°

Cosine is negative in Quadrant II.

Therefore:

cos 510° = −√3/2

Common Mistakes

Confusing sine and cosine coordinates

On the unit circle, cosine is the x-coordinate and sine is the y-coordinate.

Forgetting that the radius is 1

The unit circle is specifically defined by a radius of one unit.

Mixing degrees and radians

Always check the angle measurement being used.

Using the wrong quadrant sign

Reference angles determine magnitude, but quadrants determine signs.

Assuming sine and cosine are always positive

Both functions can have negative values.

Confusing tangent with sine or cosine

Tangent is the ratio sin θ/cos θ.

Forgetting undefined values

Tangent and secant are undefined when cosine is zero. Cosecant and cotangent are undefined when sine is zero.

Using approximate decimals when exact values are required

For example, √3/2 is an exact value, while 0.866 is an approximation.

Forgetting coterminal angles

Angles greater than 360° or less than 0° can be reduced to familiar positions.

Assuming all trigonometric functions have the same period

Sine and cosine have period 360°, while tangent has period 180°.

Incorrectly applying the Pythagorean identity

The correct identity is:

sin² θ + cos² θ = 1

not sin θ + cos θ = 1.

Key Terms

Unit Circle: A circle of radius 1 centred at the origin.

Radius: The distance from the centre of a circle to any point on its circumference.

Origin: The point (0, 0) in the Cartesian coordinate plane.

Cartesian Coordinates: Ordered pairs (x, y) describing positions in a coordinate plane.

Angle in Standard Position: An angle whose vertex is at the origin and whose initial side lies along the positive x-axis.

Initial Side: The starting ray of an angle.

Terminal Side: The ray representing the final position after rotation.

Positive Angle: An angle measured counterclockwise.

Negative Angle: An angle measured clockwise.

Degree: A unit of angular measurement in which a complete revolution contains 360°.

Radian: A unit of angular measurement based on the ratio of arc length to radius.

Coterminal Angles: Angles that share the same terminal side.

Quadrant: One of the four regions of the coordinate plane.

Reference Angle: The acute angle between an angle's terminal side and the x-axis.

Sine: The trigonometric function corresponding to the y-coordinate on the unit circle.

Cosine: The trigonometric function corresponding to the x-coordinate on the unit circle.

Tangent: The ratio of sine to cosine.

Cosecant: The reciprocal of sine.

Secant: The reciprocal of cosine.

Cotangent: The reciprocal of tangent.

Exact Trigonometric Value: A trigonometric value expressed without rounding, often using fractions and radicals.

Pythagorean Identity: The relationship sin² θ + cos² θ = 1.

Trigonometric Identity: An equation involving trigonometric functions that is true wherever both sides are defined.

Periodic Function: A function whose values repeat at regular intervals.

Period: The length of one complete repeating cycle.

Amplitude: The maximum displacement from the midline of a periodic oscillation.

Angular Frequency: The rate of change of angular position, usually measured in radians per second.

Vector Component: The part of a vector acting along a specified coordinate direction.

Check Your Understanding

1. Define the unit circle and state its equation.

2. Determine whether the point (3/5, 4/5) lies on the unit circle.

3. Explain why the coordinates of a point on the unit circle are (cos θ, sin θ).

4. Convert 150° into radians.

5. Convert 7π/6 radians into degrees.

6. State the exact coordinates of the points corresponding to 0°, 90°, 180°, and 270°.

7. Find the exact values of sin 30°, cos 45°, and tan 60°.

8. Determine the exact coordinates of the point corresponding to 120°.

9. Find sin 225° and cos 225°.

10. Evaluate tan 150°.

11. Find sin(−60°) and cos(−60°).

12. Evaluate cos 750°.

13. If sin θ = 3/5 and θ lies in Quadrant II, determine cos θ.

14. Solve sin θ = −1/2 for 0° ≤ θ < 360°.

15. A force of 80 N acts at an angle of 60° above the horizontal. Use exact trigonometric values to determine its horizontal and vertical components.

Key Takeaways

• The unit circle is a circle of radius 1 centred at the origin.

• Its equation is x² + y² = 1.

• Every point on the unit circle can be represented as (cos θ, sin θ).

• Cosine gives the x-coordinate and sine gives the y-coordinate.

• Positive angles are measured counterclockwise.

• Negative angles are measured clockwise.

• A complete revolution is 360° or 2π radians.

• Coterminal angles share the same terminal side.

• Special angles such as 30°, 45°, and 60° have exact trigonometric values.

• Reference angles allow trigonometric values to be determined in all four quadrants.

• The quadrant determines whether sine, cosine, and tangent are positive or negative.

• Tangent is calculated using tan θ = sin θ/cos θ.

• Tangent is undefined when cosine equals zero.

• The Pythagorean identity sin² θ + cos² θ = 1 follows from the equation of the unit circle.

• Sine and cosine values are always between −1 and 1.

• Sine and cosine repeat every 360° or 2π radians.

• Tangent repeats every 180° or π radians.

• The unit circle provides a geometric explanation for trigonometric graphs and identities.

• The unit circle is essential for understanding circular motion, waves, oscillations, and vector components.

• Mastering the unit circle provides the foundation for advanced trigonometry, calculus, physics, and engineering mathematics.

2. Trigonometric Functions

3. Graphs of Trigonometric Functions

4. Transformations of Trigonometric Functions

Transformations of Trigonometric Functions

Trigonometric functions can be transformed to produce a wide variety of graphs. By changing the coefficients and constants in sine, cosine, and tangent equations, we can shift graphs horizontally or vertically, reflect them, stretch or compress them, and change their periods.

These transformations are important because many real-world patterns do not follow the standard graphs of y = sin x, y = cos x, or y = tan x exactly.

For example, ocean tides may rise and fall around a particular water level, temperatures may fluctuate around an average value, and a rotating object may complete one revolution in a specified amount of time.

Transformations allow us to adjust trigonometric functions to represent these situations accurately.

The General Form of a Transformed Trigonometric Function

The general forms of transformed sine and cosine functions are:

y = A sin[B(x − h)] + D

y = A cos[B(x − h)] + D

For tangent:

y = A tan[B(x − h)] + D

Each parameter controls a particular transformation.

For sine and cosine:

Amplitude = |A|

Period = 2π/|B|

Midline: y = D

For tangent:

Period = π/|B|

Tangent does not have an amplitude because its values are unbounded.

These formulas assume B ≠ 0 and angles measured in radians.

Vertical Translations

A vertical translation moves a graph upward or downward without changing its shape.

For:

y = sin x + D

the value D determines the vertical movement.

• If D is positive, the graph moves upward.

• If D is negative, the graph moves downward.

The same principle applies to cosine and tangent.

Example: Moving a Sine Graph Upward

Consider:

y = sin x + 3

The original function is:

y = sin x

Its midline is:

y = 0

Adding 3 shifts every point upward by 3 units.

The new midline is:

y = 3

The maximum becomes:

1 + 3 = 4

The minimum becomes:

−1 + 3 = 2

The amplitude remains 1.

The period remains 2π.

Example: Moving a Cosine Graph Downward

Consider:

y = cos x − 2

The graph moves downward by 2 units.

Its midline is:

y = −2

Its maximum is:

1 − 2 = −1

Its minimum is:

−1 − 2 = −3

Important Principle

Vertical translations change the midline and output values but do not change the amplitude or period of sine and cosine.

Horizontal Translations

A horizontal translation moves a graph left or right.

For:

y = sin(x − h)

the value h determines the horizontal shift.

• If h is positive, the graph shifts right.

• If h is negative, the graph shifts left.

This can initially seem confusing because subtraction inside the function produces a shift to the right.

Example: Shift Right

Consider:

y = sin(x − π/3)

The graph shifts:

π/3 radians to the right

The original sine graph crosses its midline at x = 0 while increasing.

The transformed graph crosses its midline at:

x = π/3

while increasing.

Example: Shift Left

Consider:

y = cos(x + π/4)

Rewrite:

y = cos[x − (−π/4)]

Therefore:

h = −π/4

The graph shifts π/4 radians to the left.

Phase Shift

A horizontal translation of a periodic function is often called a phase shift.

Phase shifts are especially important when modelling waves and oscillations.

For example, two sound waves may have the same amplitude and period but reach their maximum values at different times.

Their difference can be represented by a phase shift.

Horizontal Shifts When the Coefficient B Is Present

Consider:

y = sin(2x − π)

It is tempting to say that the graph shifts π radians to the right.

However, this is incorrect.

First, factor out the coefficient of x:

y = sin[2(x − π/2)]

Now the horizontal shift is clear.

Phase shift = π/2 to the right

The coefficient B also changes the period.

Period = 2π/2

Period = π

General Rule

For an equation written as:

y = A sin(Bx + C) + D

the horizontal shift is:

h = −C/B

provided B ≠ 0.

Factoring the coefficient B from the angle expression is usually the safest approach.

Worked Example

Determine the horizontal shift of:

y = cos(3x + π/2)

Factor:

y = cos[3(x + π/6)]

Therefore:

Shift = π/6 to the left

Vertical Stretches and Compressions

The coefficient A controls the vertical scale of a sine or cosine graph.

Consider:

y = A sin x

If |A| > 1, the graph is stretched vertically.

If 0 < |A| < 1, the graph is compressed vertically.

The amplitude is:

|A|

Example: Vertical Stretch

Consider:

y = 4 sin x

The original sine graph has amplitude 1.

The transformed graph has amplitude:

4

Its maximum is 4.

Its minimum is −4.

Its period remains 2π.

Example: Vertical Compression

Consider:

y = 0.5 cos x

The amplitude is:

0.5

The graph oscillates between −0.5 and 0.5.

Vertical Scaling of Tangent

For:

y = A tan x

the coefficient A also produces a vertical stretch or compression.

However, tangent does not have an amplitude.

For example:

y = 3 tan x

has outputs three times those of y = tan x wherever defined.

Its vertical asymptotes and period remain unchanged.

Reflections Across the x-Axis

A negative coefficient A reflects a graph across the x-axis when D = 0.

For example:

y = −sin x

is the reflection of:

y = sin x

across the x-axis.

The original sine graph initially rises from the origin.

The reflected graph initially falls.

Example: Reflected Cosine

Consider:

y = −cos x

At x = 0:

y = −1

At x = π:

y = 1

The graph begins at a minimum instead of a maximum.

Reflection Across a Midline

For:

y = −3 sin x + 2

the negative coefficient reflects the sinusoidal shape across its midline:

y = 2

The amplitude is 3.

The maximum is:

2 + 3 = 5

The minimum is:

2 − 3 = −1

The midline does not change.

Important Distinction

The negative sign changes the orientation of the wave, not its amplitude.

Amplitude is always nonnegative.

Horizontal Reflections

A negative value of B produces a horizontal reflection in the basic transformation process.

For example:

y = sin(−x)

Using the odd-function identity:

sin(−x) = −sin x

Therefore:

y = sin(−x) = −sin x

A horizontal reflection of the sine graph across the y-axis is equivalent to a vertical reflection across the x-axis.

For cosine:

cos(−x) = cos x

Therefore:

y = cos(−x) = cos x

The cosine graph is unchanged because cosine is an even function.

For tangent:

tan(−x) = −tan x

These symmetry relationships can simplify transformations involving negative arguments.

Horizontal Stretches and Compressions

The coefficient B controls how quickly a trigonometric graph repeats.

For sine and cosine:

Period = 2π/|B|

For tangent:

Period = π/|B|

Example: Horizontal Compression

Consider:

y = sin(3x)

The period is:

2π/3

The graph completes three cycles over the interval from 0 to 2π.

The horizontal scale factor is 1/3.

Example: Horizontal Stretch

Consider:

y = cos(x/2)

Here:

B = 1/2

Period = 2π/(1/2)

Period = 4π

The graph takes twice as long to complete a cycle as the standard cosine graph.

Important Principle

Increasing |B| decreases the period.

Decreasing |B| toward zero increases the period.

This inverse relationship is important when interpreting frequency and repeating motion.

Amplitude, Period, and Midline

For a transformed sine or cosine function:

y = A sin[B(x − h)] + D

or:

y = A cos[B(x − h)] + D

we can identify the main graph properties directly.

| Property | Formula | |---|---| | Amplitude | |A| | | Period | 2π/|B| | | Horizontal shift | h | | Midline | y = D | | Maximum | D + |A| | | Minimum | D − |A| | | Range | [D − |A|, D + |A|] |

The maximum and minimum formulas apply even when A is negative.

Worked Example

Consider:

y = −4 cos[2(x − π/6)] + 3

Amplitude:

|−4| = 4

Period:

2π/2 = π

Horizontal shift:

π/6 to the right

Midline:

y = 3

Maximum:

3 + 4 = 7

Minimum:

3 − 4 = −1

Therefore:

• Amplitude = 4

• Period = π

• Phase shift = π/6 right

• Midline = y = 3

• Range = [−1, 7]

Sketching Transformed Sine Graphs

A transformed sine graph can be sketched using five key points over one complete cycle.

For:

y = A sin[B(x − h)] + D

with B > 0, calculate:

P = 2π/B

Then use the following x-values:

• h

• h + P/4

• h + P/2

• h + 3P/4

• h + P

The corresponding y-values are:

• D

• D + A

• D

• D − A

• D

This works for both positive and negative A.

Worked Example

Sketch one complete cycle of:

y = 2 sin(2x) + 1

Identify the Properties

Amplitude = 2

Period = 2π/2 = π

Midline: y = 1

Horizontal shift = 0

Maximum = 3

Minimum = −1

Determine Key Points

One complete cycle extends from:

x = 0 to x = π

The quarter-period is:

π/4

Plot these five points and connect them with a smooth sinusoidal curve.

The graph starts on its midline, rises to a maximum, returns to the midline, falls to a minimum, and returns to the midline.

Sketching Transformed Cosine Graphs

Cosine graphs can also be sketched using five key points.

For:

y = A cos[B(x − h)] + D

with B > 0, use the same five x-values.

The corresponding y-values are:

• D + A

• D

• D − A

• D

• D + A

Worked Example

Sketch:

y = −3 cos(x − π/2) + 2

Identify the Properties

Amplitude = 3

Period = 2π

Horizontal shift = π/2 right

Midline: y = 2

Maximum = 5

Minimum = −1

Determine Key Points

The cycle begins at:

x = π/2

The quarter-period is:

π/2

The graph begins at a minimum because A is negative.

Transformations of the Tangent Graph

Tangent transformations follow similar rules, but tangent graphs have important differences.

The general form is:

y = A tan[B(x − h)] + D

where:

• A controls vertical stretching, compression, and reflection.

• B controls the period.

• h controls the horizontal shift.

• D controls the vertical shift.

The period is:

P = π/|B|

The central point of a branch is:

(h, D)

Vertical asymptotes occur at:

x = h + (π/2 + nπ)/B

where n is any integer.

Worked Example

Sketch:

y = 2 tan[2(x − π/4)] + 1

Identify the Properties

Vertical scale factor = 2

Period:

P = π/2

Horizontal shift:

π/4 right

Midline:

y = 1

Find the Central Point

At:

x = π/4

the tangent argument is zero.

Therefore:

y = 2 tan 0 + 1

y = 1

The central point is:

(π/4, 1)

Find the Nearest Asymptotes

Half the period is:

π/4

The central branch lies between:

x = π/4 − π/4 = 0

and:

x = π/4 + π/4 = π/2

Therefore, the nearest vertical asymptotes are:

x = 0 and x = π/2

The graph increases between these asymptotes.

Writing Trigonometric Equations from Graphs

Sometimes we are given a graph and must determine an equation.

A useful approach is to identify:

• Maximum value

• Minimum value

• Amplitude

• Midline

• Period

• Horizontal shift

• Whether sine or cosine is more convenient

Worked Example

A sinusoidal graph has:

• Maximum value = 9

• Minimum value = 1

• Period = 4π

• A maximum at x = 0

Find a possible equation.

Find the Amplitude

A = (9 − 1)/2

A = 4

Find the Midline

D = (9 + 1)/2

D = 5

Find B

Period = 2π/|B|

4π = 2π/|B|

Therefore:

|B| = 1/2

Choose the Function

Because the graph has a maximum at x = 0, positive cosine is convenient.

Therefore:

y = 4 cos(x/2) + 5

This equation has the required amplitude, period, midline, and maximum.

Writing an Equation from Key Points

Suppose a sine graph:

• Crosses its midline while increasing at x = π/3.

• Has amplitude 3.

• Has period π.

• Has midline y = −2.

Find an equation.

Identify the Parameters

Amplitude:

A = 3

Period:

π = 2π/B

Therefore:

B = 2

Horizontal shift:

h = π/3

Midline:

D = −2

Write the Equation

y = 3 sin[2(x − π/3)] − 2

This equation matches all the given information.

Comparing Multiple Transformations

Consider three functions:

f(x) = sin x

g(x) = 2 sin x

h(x) = sin(2x)

Although g and h both contain the number 2, they represent different transformations.

The location of the coefficient matters.

A coefficient outside the function changes vertical values.

A coefficient multiplying x inside the function changes horizontal scaling.

Combining Transformations

A function may contain several transformations simultaneously.

Consider:

y = −2 sin[3(x + π/6)] + 4

Identify the Transformations

• Vertical stretch by factor 2.

• Reflection across the midline.

• Horizontal compression by factor 1/3.

• Shift π/6 to the left.

• Shift 4 units upward.

Determine the Properties

Amplitude:

2

Period:

2π/3

Midline:

y = 4

Maximum:

6

Minimum:

2

Determine One Cycle

The cycle begins at:

x = −π/6

The quarter-period is:

(2π/3)/4 = π/6

The graph begins at the midline and initially moves downward because the sine coefficient is negative.

Transformations in Real-World Modelling

Trigonometric transformations are particularly useful when modelling periodic systems.

Ferris Wheel Motion

A Ferris wheel has:

• Radius = 15 m

• Centre height = 18 m

• Period = 60 s

A passenger begins at the lowest point.

A suitable model is:

h(t) = 18 − 15 cos(πt/30)

where h is measured in metres and t in seconds.

Interpret the Transformations

Amplitude = 15 m

This represents the radius.

Midline = 18 m

This represents the height of the wheel's centre.

Period:

2π/(π/30) = 60 s

The negative coefficient means the passenger begins at the lowest point.

Maximum height:

18 + 15 = 33 m

Minimum height:

18 − 15 = 3 m

Find the Height After 15 Seconds

h(15) = 18 − 15 cos(π/2)

Since:

cos(π/2) = 0

h(15) = 18 m

After 15 seconds, the passenger is level with the centre of the wheel.

Modelling Ocean Tides

Suppose the water level in a harbour varies between 1 m and 5 m.

The tide completes one cycle every 12 hours.

At t = 0, the water is at its maximum height.

Find the Amplitude

A = (5 − 1)/2

A = 2 m

Find the Midline

D = (5 + 1)/2

D = 3 m

Find B

Period = 12 hours

B = 2π/12

B = π/6

Construct the Model

Because the water begins at a maximum, cosine is convenient.

h(t) = 2 cos(πt/6) + 3

Interpret

• Maximum height = 5 m

• Minimum height = 1 m

• Average level represented by the midline = 3 m

• Period = 12 hours

This simplified model assumes regular sinusoidal tides.

Actual tides may vary because of geography, weather, and multiple astronomical influences.

Modelling Temperature Changes

Seasonal temperatures can sometimes be approximated using sinusoidal functions.

Suppose the average temperature varies between 12°C and 28°C over a 12-month period.

The maximum occurs in July.

Let t represent the month number, with January corresponding to t = 1.

Determine the Amplitude

A = (28 − 12)/2

A = 8°C

Determine the Midline

D = (28 + 12)/2

D = 20°C

Determine B

Period = 12 months

B = 2π/12

B = π/6

Construct the Model

Since the maximum occurs at t = 7:

T(t) = 8 cos[(π/6)(t − 7)] + 20

This model has:

• Amplitude = 8°C

• Period = 12 months

• Midline = 20°C

• Maximum at t = 7

The horizontal shift aligns the maximum with July.

Connections to Physics and Engineering

Trigonometric transformations are essential for describing oscillating systems.

Simple Harmonic Motion

The displacement of an oscillating object may be written:

x(t) = A cos(ωt + φ)

where:

• A = amplitude

• ω = angular frequency

• φ = phase constant

• t = time

The period is:

P = 2π/ω

for positive ω.

The phase constant determines the starting position within the cycle.

Alternating Current

An alternating electrical current may be represented by:

I(t) = I₀ sin(ωt + φ)

Changing the amplitude changes the maximum current.

Changing the angular frequency changes how rapidly the current oscillates.

Changing the phase constant shifts the waveform horizontally.

Sound Waves

Sound waves can be modelled using sinusoidal functions.

• Amplitude is related to the strength of the pressure variation.

• Frequency determines pitch.

• Phase describes the position within the oscillation cycle.

A Strategy for Sketching Transformed Graphs

When sketching a transformed trigonometric function, use the following process.

Identify the parent function.

Determine whether the graph is based on sine, cosine, or tangent.

Rewrite the equation in transformation form.

For example:

y = A sin[B(x − h)] + D

Determine the vertical properties.

For sine and cosine, calculate amplitude, midline, maximum, and minimum.

Determine the period.

Use the appropriate period formula.

Determine the horizontal shift.

Identify h after factoring the coefficient B.

Identify reflections.

Check the sign of A.

Calculate key points.

Use five points for sine and cosine, or a central point and asymptotes for tangent.

Sketch the graph.

Connect sine and cosine points smoothly. Draw tangent branches between asymptotes.

Check the result.

Verify the amplitude, period, midline, and direction of the graph.

Common Mistakes

Confusing vertical and horizontal translations

A constant outside the function moves the graph vertically. A constant inside the argument changes its horizontal position.

Forgetting to factor B

For y = sin(2x − π), the shift is π/2 right, not π right.

Using A instead of |A| for amplitude

Amplitude is always nonnegative.

Using the wrong period formula

Sine and cosine use 2π/|B|. Tangent uses π/|B|.

Confusing vertical stretch with horizontal compression

A coefficient outside the function changes vertical scaling. A coefficient inside the function changes horizontal scaling.

Assuming a negative coefficient always shifts the graph

A negative A produces a reflection, not a translation.

Forgetting the midline

The graph may oscillate around a horizontal line other than y = 0.

Assuming tangent has an amplitude

Tangent is unbounded and does not have an amplitude.

Connecting tangent branches across asymptotes

Tangent graphs are discontinuous at their vertical asymptotes.

Plotting key points at incorrect intervals

For sine and cosine, consecutive key points are separated by one-quarter of the period.

Ignoring units in applications

A period might be measured in seconds, hours, days, or months.

Assuming real-world data follows a perfect sine wave

Sinusoidal models are approximations and may not capture all variations.

Key Terms

Transformation: A change to the position, shape, orientation, or scale of a graph.

Parent Function: The original function before transformations are applied.

Vertical Translation: Moving a graph upward or downward.

Horizontal Translation: Moving a graph left or right.

Phase Shift: A horizontal displacement of a periodic graph.

Vertical Stretch: Increasing vertical distances from the midline.

Vertical Compression: Decreasing vertical distances from the midline.

Horizontal Stretch: Increasing the period of a periodic graph.

Horizontal Compression: Decreasing the period of a periodic graph.

Reflection: Reversing a graph across an axis or midline.

Amplitude: The maximum vertical distance between a sinusoidal graph and its midline.

Period: The horizontal distance required for one complete cycle.

Midline: The horizontal line halfway between the maximum and minimum values.

Maximum Value: The greatest output value of a function over a specified interval.

Minimum Value: The smallest output value of a function over a specified interval.

Range: The set of possible output values.

Domain: The set of permitted input values.

Sine Function: A periodic function based on the vertical coordinate of the unit circle.

Cosine Function: A periodic function based on the horizontal coordinate of the unit circle.

Tangent Function: A periodic function defined as sine divided by cosine.

Sinusoidal Function: A function represented by a transformed sine or cosine equation.

Vertical Asymptote: A vertical line approached by a graph as its values increase or decrease without bound.

Key Point: A point used to identify the shape and position of a graph.

Angular Frequency: The rate of angular change, usually measured in radians per second.

Phase Constant: A parameter controlling the position of a periodic function within its cycle.

Periodic Model: A mathematical representation of a repeating real-world phenomenon.

Simple Harmonic Motion: Oscillatory motion that can be represented by sine or cosine functions.

Check Your Understanding

1. Explain the difference between a vertical translation and a horizontal translation.

2. Describe the transformation from y = sin x to y = sin x + 4.

3. Describe the transformation from y = cos x to y = cos(x − π/3).

4. Determine the amplitude and period of y = 5 sin(2x).

5. Find the amplitude, period, and midline of y = −3 cos x + 2.

6. Determine the horizontal shift of y = sin(4x − π).

7. Describe all transformations of y = −2 sin(x/2) − 3.

8. Find the maximum and minimum values of y = 4 cos(3x) + 1.

9. Sketch one complete cycle of y = 3 sin x + 2 using five key points.

10. Sketch one complete cycle of y = −2 cos(2x) + 1.

11. Determine the period, horizontal shift, and vertical asymptotes of y = tan[2(x − π/4)].

12. Write a sine equation with amplitude 4, period π, midline y = 3, and a rightward shift of π/6.

13. Write a cosine equation with amplitude 5, period 4π, midline y = −2, and a maximum at x = 0.

14. A Ferris wheel has a radius of 12 m, its centre is 15 m above the ground, and it completes one revolution every 48 seconds. Write a cosine model for a passenger beginning at the lowest point.

15. A periodic temperature model has a maximum of 30°C, a minimum of 10°C, a period of 24 hours, and a maximum at t = 14 hours. Write a cosine equation and explain the meaning of each parameter.

Key Takeaways

• Trigonometric transformations change the position, shape, orientation, and period of sine, cosine, and tangent graphs.

• The general transformed sine and cosine equations are y = A sin[B(x − h)] + D and y = A cos[B(x − h)] + D.

• The coefficient A controls vertical scaling and reflection.

• The amplitude of a sine or cosine graph is |A|.

• A negative A reflects the graph across its midline relative to the corresponding positive-A graph.

• The coefficient B controls horizontal scaling and period.

• The period of transformed sine and cosine functions is 2π/|B|.

• The period of transformed tangent functions is π/|B|.

• Increasing |B| decreases the period.

• Decreasing |B| toward zero increases the period.

• The parameter h controls horizontal translation.

• The parameter D controls vertical translation.

• The midline of a transformed sine or cosine graph is y = D.

• The maximum value is D + |A|.

• The minimum value is D − |A|.

• Factoring B from the angle expression helps identify the correct phase shift.

• Five key points can be used to sketch one complete cycle of a transformed sine or cosine graph.

• Tangent graphs require identification of their central points and vertical asymptotes.

• Graph features can be used to construct trigonometric equations.

• Multiple transformations can be combined within a single function.

• Transformed trigonometric functions are useful for modelling tides, Ferris wheels, seasonal temperatures, waves, and oscillations.

• Understanding trigonometric transformations provides the foundation for sinusoidal modelling, advanced trigonometry, calculus, and applications in science and engineering.

5. Inverse Trigonometric Functions

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to determine an angle when we know the value of a trigonometric ratio.

In ordinary trigonometry, we begin with an angle and calculate a ratio. In inverse trigonometry, we reverse this process: we begin with a ratio and determine the angle.

For example:

sin 30° = 1/2

Using the inverse sine function:

sin⁻¹(1/2) = 30°

Inverse trigonometric functions are essential in geometry, navigation, physics, engineering, surveying, and calculus. They allow us to calculate angles from measurements, determine directions, and solve equations involving trigonometric relationships.

What Is an Inverse Trigonometric Function?

An inverse function reverses the operation of another function.

For example, consider:

y = sin θ

The sine function takes an angle θ and produces a numerical value.

The inverse sine function reverses this relationship:

θ = sin⁻¹

This means we are finding an angle whose sine equals y.

The three principal inverse trigonometric functions are:

• Inverse sine: sin⁻¹(x), also written arcsin(x)

• Inverse cosine: cos⁻¹(x), also written arccos(x)

• Inverse tangent: tan⁻¹(x), also written arctan(x)

These functions are sometimes called arc functions.

Example

We know:

cos 60° = 1/2

Therefore:

cos⁻¹(1/2) = 60°

Similarly:

tan 45° = 1

Therefore:

tan⁻¹(1) = 45°

Important Notation

The notation sin⁻¹(x) means inverse sine.

It does not mean 1/sin(x).

These are different operations:

• sin⁻¹(x) = inverse sine

• 1/sin(x) = cosecant, written csc(x)

For example:

sin⁻¹(1/2) = 30°

but:

1/sin 30° = 2

Why Trigonometric Functions Need Restricted Domains

An inverse function must produce exactly one output for each permitted input.

However, sine, cosine, and tangent are periodic functions. They repeat their values at different angles.

For example:

sin 30° = 1/2

sin 150° = 1/2

sin 390° = 1/2

If inverse sine returned every angle with sine equal to 1/2, it would produce infinitely many answers.

To create an inverse function, we restrict the original trigonometric function to an interval where it is one-to-one.

A function is one-to-one when different inputs produce different outputs.

The chosen interval determines the principal value returned by the inverse function.

Principal Values

A principal value is the single angle returned by an inverse trigonometric function.

The principal-value intervals are:

In radians:

Notice that inverse tangent never returns exactly −90° or 90°.

These angles are excluded because tangent is undefined there.

The Inverse Sine Function

The inverse sine function is:

y = sin⁻¹(x)

It answers the question:

What angle has a sine equal to x?

The domain is:

−1 ≤ x ≤ 1

The range is:

−π/2 ≤ y ≤ π/2

or:

−90° ≤ y ≤ 90°

Common Exact Values

Worked Example

Evaluate:

sin⁻¹(√3/2)

We know:

sin 60° = √3/2

Since 60° lies in the principal-value range:

sin⁻¹(√3/2) = 60° = π/3

Negative Values

Evaluate:

sin⁻¹(−1/2)

We know:

sin(−30°) = −1/2

Therefore:

sin⁻¹(−1/2) = −30° = −π/6

Domain Restrictions

The expression:

sin⁻¹(2)

is undefined over the real numbers because sine cannot produce a value greater than 1.

The Inverse Cosine Function

The inverse cosine function is:

y = cos⁻¹(x)

It answers:

What angle has a cosine equal to x?

Its domain is:

−1 ≤ x ≤ 1

Its range is:

0 ≤ y ≤ π

or:

0° ≤ y ≤ 180°

Common Exact Values

Worked Example

Evaluate:

cos⁻¹(−1/2)

We know:

cos 120° = −1/2

Therefore:

cos⁻¹(−1/2) = 120° = 2π/3

Why the Answer Is Not 240°

Cosine is also −1/2 at 240°.

However, inverse cosine returns only angles between 0° and 180°.

Therefore, 120° is the principal value.

The Inverse Tangent Function

The inverse tangent function is:

y = tan⁻¹(x)

It answers:

What angle has a tangent equal to x?

Its domain is:

All real numbers

Its range is:

−π/2 < y < π/2

or:

−90° < y < 90°

Common Exact Values

Worked Example

Evaluate:

tan⁻¹(√3)

Since:

tan 60° = √3

Therefore:

tan⁻¹(√3) = 60° = π/3

Worked Example

Evaluate:

tan⁻¹(−1)

Since:

tan(−45°) = −1

Therefore:

tan⁻¹(−1) = −45° = −π/4

Comparing the Three Principal Inverse Functions

The inverse sine and inverse cosine functions have restricted domains because sine and cosine values cannot exceed 1 or fall below −1.

Inverse tangent has no such domain restriction because tangent can produce any real value.

Interpreting Inverse Functions Graphically

The graph of an inverse function is the reflection of the original function's restricted graph across the line:

y = x

This happens because the input and output values exchange roles.

If the original function contains a point:

(a, b)

then its inverse contains:

(b, a)

Example: Sine and Inverse Sine

The sine function contains:

(π/6, 1/2)

The inverse sine function contains:

(1/2, π/6)

The coordinates are reversed.

Example: Cosine and Inverse Cosine

The cosine function contains:

(π/3, 1/2)

The inverse cosine function contains:

(1/2, π/3)

Again, the coordinates are reversed.

Example: Tangent and Inverse Tangent

The tangent function contains:

(π/4, 1)

The inverse tangent function contains:

(1, π/4)

The graphs reflect across y = x when tangent is restricted to its principal interval.

Important Graphical Features

Inverse Sine

The graph:

• Begins at (−1, −π/2).

• Passes through (0, 0).

• Ends at (1, π/2).

• Increases continuously.

Inverse Cosine

The graph:

• Begins at (−1, π).

• Passes through (0, π/2).

• Ends at (1, 0).

• Decreases continuously.

Inverse Tangent

The graph:

• Passes through (0, 0).

• Increases continuously.

• Approaches y = π/2 as x increases without bound.

• Approaches y = −π/2 as x decreases without bound.

The horizontal asymptotes are never reached.

Evaluating Inverse Trigonometric Expressions

Inverse trigonometric expressions may involve fractions, radicals, and combinations of functions.

Worked Example: sin⁻¹(1/2) + cos⁻¹(1/2)

Evaluate each function.

sin⁻¹(1/2) = π/6

cos⁻¹(1/2) = π/3

Add:

π/6 + π/3

= π/6 + 2π/6

= π/2

Worked Example: 2tan⁻¹(1)

tan⁻¹(1) = π/4

Therefore:

2tan⁻¹(1) = 2(π/4)

= π/2

Worked Example: cos⁻¹(−√2/2) − sin⁻¹(1/2)

cos⁻¹(−√2/2) = 3π/4

sin⁻¹(1/2) = π/6

Subtract:

3π/4 − π/6

= 9π/12 − 2π/12

= 7π/12

Compositions of Trigonometric and Inverse Functions

A composition involves applying one function to the output of another.

For example:

sin(sin⁻¹ x) = x

for −1 ≤ x ≤ 1.

Similarly:

cos(cos⁻¹ x) = x

for −1 ≤ x ≤ 1.

And:

tan(tan⁻¹ x) = x

for every real x.

These identities work because the inverse function reverses the original operation.

Worked Example

Evaluate:

sin[sin⁻¹(3/5)]

The inverse sine produces an angle whose sine is 3/5.

Applying sine returns the original value.

Answer = 3/5

Important Restriction

The reverse composition requires more care.

For example:

sin⁻¹(sin 150°)

We know:

sin 150° = 1/2

Therefore:

sin⁻¹(1/2) = 30°

sin⁻¹(sin 150°) = 30°

It is not 150° because inverse sine must return a principal value.

Another Example

Evaluate:

cos⁻¹(cos 240°)

cos 240° = −1/2

Therefore:

cos⁻¹(−1/2) = 120°

Answer = 120°

Using Right Triangles to Evaluate Compositions

Inverse trigonometric functions can be combined with other trigonometric functions.

For example:

cos[sin⁻¹(3/5)]

Let:

θ = sin⁻¹(3/5)

Then:

sin θ = 3/5

Since θ lies in the principal range of inverse sine and sine is positive, θ is in Quadrant I.

Construct a right triangle with:

• Opposite side = 3

• Hypotenuse = 5

Using the Pythagorean theorem:

Adjacent² = 5² − 3²

Adjacent² = 25 − 9

Adjacent = 4

Therefore:

cos θ = adjacent/hypotenuse

cos[sin⁻¹(3/5)] = 4/5

Worked Example

Evaluate:

sin[tan⁻¹(2)]

Let:

θ = tan⁻¹(2)

Then:

tan θ = 2 = 2/1

Construct a right triangle with:

• Opposite = 2

• Adjacent = 1

Find the hypotenuse:

r = √(2² + 1²)

r = √5

Therefore:

sin θ = 2/√5

Rationalise:

sin[tan⁻¹(2)] = 2√5/5

Solving Equations with Inverse Trigonometric Functions

Inverse trigonometric functions can be used to solve equations in which an unknown angle appears inside a trigonometric function.

Worked Example: sin θ = 0.6

Find the acute angle θ.

Apply inverse sine:

θ = sin⁻¹(0.6)

Using a calculator in degree mode:

θ ≈ 36.87°

Worked Example: cos θ = 0.25

For an acute angle:

θ = cos⁻¹(0.25)

θ ≈ 75.52°

Worked Example: tan θ = 1.5

For an acute angle:

θ = tan⁻¹(1.5)

θ ≈ 56.31°

These calculations are common in right-triangle problems.

Finding All Solutions to Trigonometric Equations

An inverse trigonometric function returns one principal angle.

However, a trigonometric equation may have several solutions over a specified interval.

Worked Example

Solve:

sin θ = 1/2

for:

0° ≤ θ < 360°

First find the principal angle:

θ = sin⁻¹(1/2)

θ = 30°

Sine is positive in Quadrants I and II.

The second angle is:

180° − 30° = 150°

Therefore:

θ = 30° or 150°

Worked Example

Solve:

cos θ = −1/2

for:

0° ≤ θ < 360°

The principal inverse cosine value is:

cos⁻¹(−1/2) = 120°

Cosine is negative in Quadrants II and III.

The second solution is:

360° − 120° = 240°

Therefore:

θ = 120° or 240°

Worked Example

Solve:

tan θ = 1

for:

0° ≤ θ < 360°

The principal angle is:

tan⁻¹(1) = 45°

Tangent repeats every 180°.

Therefore:

45° + 180° = 225°

θ = 45° or 225°

Solving Equations Containing Inverse Functions

Sometimes the inverse function itself appears in an equation.

Worked Example

Solve:

sin⁻¹(x) = π/6

Apply sine to both sides:

x = sin(π/6)

x = 1/2

This satisfies the domain of inverse sine.

Worked Example

Solve:

2cos⁻¹(x) = 2π/3

Divide by 2:

cos⁻¹(x) = π/3

Apply cosine:

x = cos(π/3)

x = 1/2

Worked Example

Solve:

tan⁻¹(2x − 1) = π/4

Apply tangent:

2x − 1 = tan(π/4)

2x − 1 = 1

2x = 2

x = 1

Inverse Trigonometry in Right-Triangle Problems

One of the most common applications of inverse trigonometric functions is finding unknown angles in right triangles.

Choosing the Correct Function

Use inverse sine when the opposite side and hypotenuse are known.

θ = sin⁻¹(opposite/hypotenuse)

Use inverse cosine when the adjacent side and hypotenuse are known.

θ = cos⁻¹(adjacent/hypotenuse)

Use inverse tangent when the opposite and adjacent sides are known.

θ = tan⁻¹(opposite/adjacent)

Worked Example: Finding an Angle

A right triangle has:

• Opposite side = 8 cm

• Adjacent side = 15 cm

Find θ.

Use tangent:

tan θ = 8/15

Therefore:

θ = tan⁻¹(8/15)

θ ≈ 28.07°

The unknown angle is approximately 28.1°.

Application: Angle of Elevation

An observer stands 40 m from the base of a building.

The top of the building is 30 m above the observer's eye level.

Find the angle of elevation.

Identify the Triangle

Opposite = 30 m

Adjacent = 40 m

Use tangent:

tan θ = 30/40

tan θ = 0.75

Apply inverse tangent:

θ = tan⁻¹(0.75)

θ ≈ 36.87°

The angle of elevation is approximately 36.9°.

Real-World Connection

Surveyors and engineers use similar calculations to determine heights and angles without directly measuring inaccessible distances.

Application: Navigation

A ship travels 12 km east and 5 km north.

Determine its direction relative to east.

Identify the Components

Horizontal displacement = 12 km

Vertical displacement = 5 km

The direction angle satisfies:

tan θ = 5/12

Therefore:

θ = tan⁻¹(5/12)

θ ≈ 22.62°

The ship's displacement is directed approximately:

22.6° north of east

The total displacement is:

d = √(12² + 5²)

d = √169

d = 13 km

Important Note About Direction

The ordinary inverse tangent function returns angles between −90° and 90°.

When finding a direction from horizontal and vertical components, the signs of both components must be considered to determine the correct quadrant.

For example, a vector pointing southwest requires an angle in Quadrant III, even though tan⁻¹(y/x) alone may return a positive acute angle.

In computing, the function often called atan2(y, x) accounts for both coordinate signs and determines the appropriate direction.

Application: Inclined Planes

A ramp rises 1.5 m over a horizontal distance of 6 m.

Find the angle the ramp makes with the horizontal.

Identify the Ratio

Opposite = 1.5 m

Adjacent = 6 m

Therefore:

tan θ = 1.5/6

tan θ = 0.25

Apply inverse tangent:

θ = tan⁻¹(0.25)

θ ≈ 14.04°

The ramp makes an angle of approximately 14° with the horizontal.

This calculation is useful in architecture, accessibility design, and civil engineering.

Application: Forces and Vectors

A force has horizontal and vertical components:

Fₓ = 24 N

Fᵧ = 10 N

Determine the magnitude and direction of the force.

Find the Magnitude

F = √(Fₓ² + Fᵧ²)

F = √(24² + 10²)

F = √676

F = 26 N

Find the Direction

tan θ = Fᵧ/Fₓ

tan θ = 10/24

θ = tan⁻¹(10/24)

θ ≈ 22.62°

The force has magnitude 26 N and acts approximately 22.6° above the positive horizontal direction.

This method is widely used in mechanics, engineering, and vector analysis.

Application: Calculus

Inverse trigonometric functions also appear in differentiation and integration.

Their derivatives include:

d/dx[arcsin(x)] = 1/√(1 − x²)

for −1 < x < 1.

d/dx[arccos(x)] = −1/√(1 − x²)

for −1 < x < 1.

d/dx[arctan(x)] = 1/(1 + x²)

for all real x.

These formulas assume angles are measured in radians.

Example: Differentiating Inverse Tangent

Let:

f(x) = arctan(3x)

Using the chain rule:

f′(x) = 3/(1 + (3x)²)

Therefore:

f′(x) = 3/(1 + 9x²)

Connection to Integration

Inverse tangent appears when integrating expressions such as:

∫ 1/(1 + x²) dx

The result is:

arctan(x) + C

Inverse trigonometric functions are therefore important in advanced integration techniques.

Using Calculators Correctly

Most scientific calculators include inverse trigonometric functions.

They are commonly accessed using a secondary function key.

For example:

• sin⁻¹

• cos⁻¹

• tan⁻¹

Degree Mode

If a question asks for an angle in degrees, use degree mode.

For example:

tan⁻¹(1) = 45°

Radian Mode

If a question asks for radians, use radian mode.

For example:

tan⁻¹(1) = π/4

Approximately:

0.7854 radians

Checking Results

A useful way to verify an inverse trigonometric calculation is to apply the original trigonometric function.

For example:

θ = sin⁻¹(0.6) ≈ 36.87°

Check:

sin 36.87° ≈ 0.6

The result is consistent.

A Strategy for Inverse Trigonometric Problems

When solving a problem involving inverse trigonometric functions:

1. Identify the known trigonometric ratio or expression.

2. Determine which inverse function is appropriate.

3. Check that the input lies within the function's domain.

4. Find the principal value.

5. Determine whether additional angles are required.

6. Use quadrant information when necessary.

7. Convert between degrees and radians if required.

8. Check the result using the original trigonometric function.

9. Include units and interpret the result in context.

This process helps prevent errors involving principal values and quadrant restrictions.

Common Mistakes

Confusing inverse functions with reciprocals

sin⁻¹(x) is inverse sine, while 1/sin(x) is cosecant.

Ignoring principal-value ranges

Inverse trigonometric functions return values only within their defined principal intervals.

Assuming inverse sine returns every possible angle

Inverse sine returns one principal angle, not all solutions to a trigonometric equation.

Forgetting additional solutions

An equation such as sin θ = 1/2 has multiple solutions over a full revolution.

Using an input outside the domain

For real-valued inverse sine and inverse cosine, the input must be between −1 and 1.

Confusing degrees and radians

Calculator settings must match the units required by the problem.

Assuming sin⁻¹(sin θ) always equals θ

This is true only when θ lies within the principal-value range of inverse sine.

Ignoring quadrant information in vector problems

Inverse tangent alone may not determine the correct direction.

Using the wrong trigonometric ratio

Choose sine, cosine, or tangent based on the known sides of the right triangle.

Rounding too early

Keep sufficient precision during calculations and round only the final answer.

Forgetting units

Practical problems may require angles in degrees or radians and distances in metres or kilometres.

Key Terms

Inverse Function: A function that reverses the relationship between the inputs and outputs of another function.

Inverse Trigonometric Function: A function that determines an angle from a trigonometric ratio.

Inverse Sine: The function arcsin(x), which returns an angle whose sine equals x.

Inverse Cosine: The function arccos(x), which returns an angle whose cosine equals x.

Inverse Tangent: The function arctan(x), which returns an angle whose tangent equals x.

Principal Value: The unique angle returned by an inverse trigonometric function within its specified range.

Restricted Domain: A limited interval of input values chosen to make a function one-to-one.

One-to-One Function: A function in which different inputs produce different outputs.

Domain: The set of permitted input values.

Range: The set of possible output values.

Periodic Function: A function whose values repeat at regular intervals.

Composition of Functions: Applying one function to the output of another.

Inverse Relationship: A relationship in which two functions reverse each other's operations on appropriate domains.

Reflection Across y = x: The geometric relationship between the graph of a one-to-one function and its inverse.

Horizontal Asymptote: A horizontal line that a graph approaches as the input increases or decreases without bound.

Exact Value: A value expressed without rounding, often using fractions, radicals, or π.

Radian: A unit of angular measurement based on the ratio of arc length to radius.

Degree: A unit of angular measurement in which a complete revolution contains 360°.

Angle of Elevation: The angle measured upward from the horizontal to an object.

Angle of Depression: The angle measured downward from the horizontal to an object.

Vector Component: The part of a vector acting along a particular coordinate direction.

Direction Angle: An angle describing the orientation of a vector.

Quadrant: One of the four regions of the Cartesian coordinate plane.

Arctangent2: A two-input angle calculation that uses horizontal and vertical components to determine the correct quadrant.

Chain Rule: A differentiation rule used for composite functions.

Antiderivative: A function whose derivative equals a given function.

Check Your Understanding

1. Define an inverse trigonometric function and explain how it differs from a reciprocal trigonometric function.

2. State the domains and principal-value ranges of arcsin(x), arccos(x), and arctan(x).

3. Evaluate sin⁻¹(1/2) in degrees and radians.

4. Evaluate cos⁻¹(−√3/2).

5. Evaluate tan⁻¹(−1).

6. Explain why sin⁻¹(2) is undefined over the real numbers.

7. Evaluate sin⁻¹(sin 150°).

8. Evaluate cos⁻¹(cos 240°).

9. Simplify cos[sin⁻¹(5/13)].

10. Solve 2tan⁻¹(x) = π/2.

11. Solve sin θ = √3/2 for 0° ≤ θ < 360°.

12. Solve cos θ = −√2/2 for 0° ≤ θ < 360°.

13. A ladder reaches 6 m up a wall and its base is 2.5 m from the wall. Calculate the angle the ladder makes with the ground.

14. A displacement vector has components 8 km east and 15 km south. Determine its magnitude and direction relative to east.

15. A drone travels 120 m horizontally and rises 45 m vertically. Determine its angle of elevation and total displacement. Explain which inverse trigonometric function you used and why.

Key Takeaways

• Inverse trigonometric functions determine angles from known trigonometric ratios.

• The three principal inverse functions are arcsin(x), arccos(x), and arctan(x).

• Inverse trigonometric functions are different from reciprocal trigonometric functions.

• Trigonometric functions require restricted domains to have well-defined inverse functions.

• A principal value is the unique angle returned by an inverse trigonometric function.

• The domain of arcsin(x) is [−1, 1], and its range is [−π/2, π/2].

• The domain of arccos(x) is [−1, 1], and its range is [0, π].

• The domain of arctan(x) is all real numbers, and its range is (−π/2, π/2).

• Inverse sine and inverse tangent are increasing functions.

• Inverse cosine is a decreasing function.

• The graphs of inverse functions are reflections of their restricted original functions across y = x.

• The inverse tangent graph has horizontal asymptotes at y = ±π/2.

• Applying a trigonometric function to its corresponding inverse returns the original input when that input is in the inverse function's domain.

• Applying an inverse trigonometric function to a trigonometric function may return a principal angle different from the original angle.

• Solving trigonometric equations may require finding additional solutions beyond the principal value.

• Inverse trigonometric functions are useful for calculating unknown angles in right triangles.

• Inverse tangent is commonly used to calculate angles of elevation, slopes, and vector directions.

• Quadrant information is essential when determining directions from vector components.

• Calculator angle settings must match the required units.

• Inverse trigonometric functions also appear in differentiation and integration.

• Understanding inverse trigonometric functions provides a foundation for solving geometric, scientific, engineering, and advanced calculus problems involving angles and directions.